(II) A diverging lens with a focal length of -14cm is placed 1...

Question

(II) A diverging lens with a focal length of -14cm is placed 12 cm to the right of a converging lens with a focal length of 18 cm. An object is placed 29 cm to the left of the converging lens. (a) Where will the final image be located? (b) Where will the image be if the diverging lens is 38 cm from the converging lens?

Accepted Answer

Welcome back. Everyone in this problem. A two lens system consists of a converging lens with a focal length 20 centimeters and a diverging lens with a focal length of 12 centimeters. The diverging lens is placed 15 centimeters to the right of the converging lens with an object 30 centimeters to the left of the converging lens. If the diverging lens is moved to 40 centimeters from the converging lens, where will the final image be relative to the diverging lens? In both setups? A says it's 16 centimeters to the left in the first set up and 30 centimeters to the left in the second set up. B says it's 30 centimeters to the left and 16 centimeters to the left respectively. C 40 centimeters to the left and 14 centimeters to the left and D 60 centimeters to the left and 11 centimeters to the left. Now, we have quite a few lengths here. So let's break them down step by step while we try to find where the final image will be relative to the diverging lens in both setups. Now let's first start by trying to find the image formed by the converging lens. Recall by the lens equation OK. That one divided by the object distance from the converging lens D plus one divided by D OK, which is the distance the image distance sorry from the converging lens is going to be equal to one divided by F the focal length of the converging lens. Now, in this case, we know that our object distance from the converging lens is 30 centimeters. And we know that the focal length, as our problem tells us for our converging length is 20 centimeters. So with that in mind, then we can rearrange our formula to solve for our image distance or I the distance of the image formed or from the converging lens. OK? Because no, this tells us that the reciprocal of D I then equals one divided by the focal length minus one divided by D. OK. And now if we combine and simplify, then di is going to be equal to dof divided by do minus F if we substitute these values that 30 centimeters multiplied by 20 centimeters divided by 30 minus 20 centimeters. And now when we go ahead and solve even further, we get D I to be 60 centimeters. So what does this mean? Well, the image formed by the converging lens is going to be 60 centimeters to the right of the converging lens. Now that we have this information, then we can use it to find the image found by the divergent lens in both setups let's start with the first set up. If we go back to our problem, it tells us that in the first set up, let me highlight that in blue, the diverging lens is placed 15 centimeters to the right of the converging lens with an object 30 centimeters to the left of the converging lens. OK. So now Matt tells us then in the first setup, OK, that the image formed by the converging lens, which is now acting as the object for the diverging lens and the distance between the two lenses is 15 centimeters. So the object distance for the diverging lens, let's call that D two is going to be equal to the negative value of 60 centimeters minus 15 centimeters which equals negative 45 centimeters. Now, we can use the lens equation for the diverging lens because with that idea, OK, then we can say that one divided by do two minus one, divided by D I two just to distinguish this from our D and D I from the converging lens is going to be equal to one divided by F two. And now we know that F two is the focal length of the diverging lens negative 12 centimeters while we just sold for the two to be negative 45 centimeters. Now we can do some transposition here to solve ad two similar to what we had in our first formula because now if we use the same idea but with different variables. This tells us that DI two OK is gonna be equal to D two, F two minus D two minus F two or sorry, divided by sorry do two minus F two. Now we can substitute these values, these known values that's negative 45 centimeters multiplied by negative 12 centimeters all divided by negative 45 centimeters minus negative 12 centimeters. And now when we calculate here and we rounded to four significant figures, actually, then we should get the distance from our diverging lens in the initial set up to be equal to approximately negative 16.36 centimeters or to two significant figures, approximately 16, negative 16 centimeters and no negative denotes that it's to the left. Therefore, the final image is 16 centimeters to the left of the diverging lens in our first set up. Now, what about our second set up? Let's go back to our problem. It says that in our second set up, the diverging lens is moved 40 centimeters from the converging lens. Now, the distance between the lenses, the lenses changes to 40 centimeters. So we're going to use the same idea from the first set up. But now we know that that distance between the lenses is 40 centimeters instead of 15 centimeters. So now in the second set up, let me put that in red. OK, we know that the two is gonna be the negative value of 60 centimeters minus 40 centimeters, which is negative 20 centimeters. And now using the lens equation again, for the diverging lens, that means di two, OK. Is gonna be equal to D two M sorry, do two multiplied by F two divided by D two minus F two. OK. Which is gonna be equal to negative 20 centimeters multiplied by negative 12 centimeters, same focal length divided by negative 20 centimeters minus negative 12 centimeters. I know when we work that out, we get our value to be negative 30 centimeters OK, or 30 centimeters to the left. Therefore, when the diverging lenses move to 40 centimeters from the converging lens, the final image is 30 centimeters to the left of the diverging lens. Whereas in the first set up, the final image is 16 centimeters to the left of the diverging lens. If we look back at our answer choices, then that tells us a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Lens Formula

Sign Convention for Lenses

Image Formation by Lenses

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